Problem 49 - A Course in Calculus and Real Analysis

Question

Let \(f:[a, b] \rightarrow \mathbb{R}\) be a monotonic function. If \(G:[a, b] \rightarrow \mathbb{R}\) is differentiable and \(G^{\prime}\) is continuous, then show that there is \(c \in[a, b]\) such that $$ \int_{a}^{b} f(x) G^{\prime}(x) d x=f(b) G(b)-f(a) G(a)-G(c)[f(b)-f(a)] $$ (Hint: Given any partition \(P=\left\\{x_{0}, x_{1}, \ldots, x_{n}\right\\}\) of \([a, b]\), consider the sum \(\sum_{i=1}^{n} f\left(x_{i}\right)\left[G\left(x_{i}\right)-G\left(x_{i-1}\right)\right] .\) Write it as \(f(b) G(b)-f(a) G(a)-\) \(\sum_{i=1}^{n} G\left(x_{i-1}\right)\left[f\left(x_{i}\right)-f\left(x_{i-1}\right)\right]\) and also as \(\sum_{i=1}^{n} f\left(x_{i}\right) G^{\prime}\left(s_{i}\right)\left(x_{i}-x_{i-1}\right)\) for some \(s_{i} \in\left[x_{i-1}, x_{i}\right] .\) Use the Theorem of Bliss (Exercise 48) and the inequalities \(m(g)[f(b)-f(a)] \leq \sum_{i=1}^{n} G\left(x_{i-1}\right)\left[f\left(x_{i}\right)-f\left(x_{i-1}\right)\right] \leq\) \(M(g)[f(b)-f(a)] .)\)

Step-by-Step Solution

Answer

In summary, given a monotonic function \(f\) and a differentiable function \(G\) with continuous derivative \(G'\), we can show that there exists a \(c \in [a, b]\) satisfying the equation: $$ \int_{a}^{b} f(x) G^{\prime}(x) d x = f(b)G(b) - f(a)G(a) - G(c)[f(b) - f(a)] $$ To prove this, we use the given partition \(P\), Theorem of Bliss (Integration by Parts), and the provided inequalities. After comparing the result with the expression for \(S\), we can conclude that the desired equation holds for some value of \(c\).

Step 1: Write the sum using partition P

We are given a partition \(P = \{x_0, x_1, ..., x_n\}\) of \([a, b]\). We can use it to consider the sum: $$S=\sum_{i=1}^{n} f(x_i)[G(x_i)-G(x_{i-1})]$$ We can rewrite it to get: $$S = f(b)G(b) - f(a)G(a) - \sum_{i=1}^n G(x_{i-1})[f(x_i) - f(x_{i-1})]$$

Step 2: Using Theorem of Bliss

According to the theorem of integration by parts (Theorem of Bliss), we can rewrite the sum as follows: $$S = \int_{a}^{b} f(x)G'(s_i)(x_i - x_{i-1})dx$$ for some \(s_i \in [x_{i-1}, x_i]\).

Step 3: Finding c using the inequalities

We have the inequalities provided as: $$m(g)[f(b)-f(a)] \leq \sum_{i=1}^{n} G(x_{i-1})[f(x_{i})-f(x_{i-1})] \leq M(g)[f(b)-f(a)]$$ Using these inequalities, we know $$ m(g)[f(b)-f(a)] \leq \int_{a}^{b} f(x)G'(s_i)(x_i-x_{i-1})dx \leq M(g)[f(b)-f(a)] $$ Since \(G'(s_i)\) is continuous and strictly between \(m(g)\) and \(M(g)\), there exists a \(c \in [a, b]\) such that: $$ \int_{a}^{b} f(x)G'(x)dx = G(c)[f(b)-f(a)] $$ Comparing this equation with the equation for \(S\) from step 1, we have: $$ \int_{a}^{b} f(x) G^{\prime}(x) d x = f(b)G(b)-f(a)G(a)-G(c)[f(b)-f(a)] $$ Therefore, we can conclude there exists a \(c\in[a, b]\) where the given equation holds.

Key Concepts

Monotonic Functions

Monotonic functions are a fundamental concept used to describe functions that either never decrease or never increase as their input value increases. They come in two forms:

• Monotonically Increasing: A function that satisfies \(f(x_1) \leq f(x_2)\) for all \(x_1 < x_2\) within a specific interval.
• Monotonically Decreasing: A function that maintains \(f(x_1) \geq f(x_2)\) for all \(x_1 < x_2\).

Monotonic functions are particularly useful in mathematical analysis due to their predictable behavior. In our problem, the given function \(f:[a, b] \rightarrow \mathbb{R}\) is monotonic. This allows us to employ specific theorems and properties, ensuring our solutions or calculations remain valid within this interval. These characteristics also support simple and complex integrals, as monotonic functions do not have oscillations that might complicate the computation.

Continuous Functions

Continuous functions are those that have no breaks, jumps, or gaps in their graphs, which means you can draw them without lifting your pen. Formally, a function is continuous at a point \(c\) if:

• \(\lim_{x \to c} f(x) = f(c)\)
• The limit exists and equals the function's value at that point.

In broader terms, a function is continuous in an interval if it is continuous at every point within that interval. For the discussed problem, continuity is significant when we consider \(G':[a, b] \rightarrow \mathbb{R}\) continuous. This continuity assures us that the Fundamental Theorem of Calculus can be applied, which in turn allows the manipulation and estimation of integrals, critical for finding the specific constant \(c\).

Differentiable Functions

Differentiable functions are those that have a derivative at every point within their domain. This implies that the function is smooth, with a well-defined tangent at every point, which suggests no sharp corners or cusps. When we have a differentiable function, \(G:[a, b] \rightarrow \mathbb{R}\), it means:

• \(G\) is continuous and also possesses derivative \(G'\) over the entire interval \([a,b]\).
• This differentiability means we can directly use integration by parts, which integrates a product of two functions by combining their derivatives and antiderivatives in a useful way.

In our case, the differentiability of \(G\) and continuity of \(G'\) play vital roles in applying the Theorem of Bliss, leading to the discovery of the constant \(c\) that satisfies the integral equation given in the exercise.

Theorem of Bliss

The Theorem of Bliss is a pivotal result that builds on the classical technique of integration by parts. It asserts that for functions that are monotonic, integrable, and have certain continuity conditions, specific integral comparisons can be effectively made.

• It is useful in converting the integration of a product of functions into an evaluation at endpoints plus additional integral terms.
• In the exercise, using the Theorem of Bliss allowed expressing the sum \(S\) in different forms and led to understanding how the integral and the inequalities involving \(m(g)\) and \(M(g)\) relate to each other.

The theorem signals that even if two functions complicate direct integration, under specific conditions (such as monotonicity or continuity), their behavior over an interval can still be reliably analyzed. This allows us to find an answer to the integral equation from the problem by structuring it in a way reliant on the endpoints and some constant \(c\).